# Zone of negative curvature on surfaces embedded in $\mathbb{R}^3$

I consider the standard embedding of a compact oriented surface $\Sigma$ (say of genus 2) in the Euclidean space $\mathbb{R}^3$. I have coloured on the picture below the zone of this surface where the curvature is negative (in blue) and where it is positive (in red).

This produces a quite remarkable partition of the surface in the sense that it distinguishes two (homotopy classes) of simple closed curves.

My philosophical question is: to which extend are these simple closed curves characteristic of the embedding? are these curves going to survive deformations through embeddings?

As it is easily seen, one could make this picture more complicated by pinching closed curves and give birth to new non-homotopically trivial zone of negative curvature.

However, this operation seems quite artificial. So would be the operation consisting in adding little bumps within the blue zone to create a little island of red surrounded by a blue zone.

So a more precise question is probably: is there a way of deforming this embedding in $\mathbb{R}^3$ that destroys the blue zone (of the first picture) in an essential way? also, is it possible that the fundamental group of the negative curvature locus embeds trivially (by that I mean that its fundamental group is killed by the inclusion map in the surface) after deformation of this initial example?

Thanks!

• I like this question, and I also feel a need to mention that my girlfriend saw me reading it and asked me 'if i was studying the mathematics of spiderman masks' -- indirectly, I suppose we are. – Alec Rhea Jul 10 '17 at 22:10
• I am not sure I get what you mean by "embedding trivially", could you precise? Do you ask whether it is possible that the blue region has trivial fundamental group (then talking about its embedding is a bit odd), that the blue region has a fundamental group that is killed by the inclusion map in the surface (in which case it is not embedded), or something else? – Benoît Kloeckner Jul 11 '17 at 12:27
• I mean 'the blue region has a fundamental group that is killed by the inclusion map in the surface'. I edit it right away. – Selim G Jul 11 '17 at 12:51

You can't destroy the blue zone in the first picture entirely because then the minimizing closed geodesic in the implied free homotopy class would admit of a shortening in the direction of the normal vector (if the Gaussian curvature is positive along the curve then the curve could be shortened by second variation formula).

• I don't get your argument exactly. What is then the shortest deformation of the outermost geodesic in the first picture of the question? – Benoît Kloeckner Jul 10 '17 at 19:15
• Perhaps the point is that the two blue geodesics are stable? – Igor Rivin Jul 10 '17 at 21:18
• @BenoîtKloeckner, I agree with you that the question is ambiguous and therefore it is not really clear what my answer answers exactly. If you think this answer is misleading let me know and I will delete it. – Mikhail Katz Jul 11 '17 at 9:32
• @IgorRivin, I think that's the right viewpoint: that there must be some negative curvature around the homotopy/homology systole because if it was all positive you would easily deform it to a shorter curve in the same homotopy/homology class. I wonder if there is a way to quantify the "some negative curvature". – alvarezpaiva Jul 12 '17 at 7:42

You can deform the embedding into an homotopically distinct embedding with the same image, thus changing the homotopy classes of curves that correspond to blue regions. Indeed, you can deform your embedding to a more symmetric one: two concentric spheres, one slightly inside the other, with three "collars" between them (each collar being obtained by removing two disks, one in each sphere, and replacing them with an annulus), regularly positioned. Then by a one third of turn rotation and the inverse deformation, you obtain an embedding with the same image as before, but with one of the previously blue curve sent to the outer most curve.

Added in edit: to see why my claim is true, it is easier to look at the deformation from the symmetric position to the one pictured in the question. For this, simply enlarge one of the collars, turn the embedding so that the enlarged collar is on top, then push everything down. Hope this is clearer. In case not, let me quote the exercise that made me think about this: show that the boundary of a small tubular neighborhood of the $1$-skeleton of a icosahedron is a surface of genus 19.

• Benoit, if I understood your answer correctly, you are moving the blue region elsewhere rather than getting rid of it! – Mikhail Katz Jul 10 '17 at 15:29
• Hi! It is not obvious to me that I can deform the original embedding into yours. Should it be? Also, I have the impression that at some point (when you do the turn by one third and the inverse map) you are moving from a continuous motion to precomposing the embedding by a diffeomorphism of $\Sigma$ which is not isotopic to the identity, right? – Selim G Jul 10 '17 at 15:37
• @MikhailKatz: you might be right, but it mostly comes down on what the question precisely is. Also, in the process some blue region disappears at some point (in the symmetric position, there are three blue regions). – Benoît Kloeckner Jul 10 '17 at 19:03

I believe this circle of questions (on how "stable" stable geodesics are, and what that means for mankind) is answered in this paper (google the title, and the paper is available for free).

Zhang, Min, et al. "Stable geodesic surface signatures." Tsinghua Science and Technology 17.4 (2012): 471-480.